Problem: $f(x, y, z) = (x\cos(z), y\sin(z), z\tan(z))$ $\text{curl}(f) = $ $\hat{\imath} + $ $\hat{\jmath} + $ $\hat{k}$
Answer: $f(x, y, z) = (f_0, f_1, f_2)$ The curl of $f$ : $\begin{aligned} \text{curl}(f) &= \det \begin{bmatrix} {\hat{\imath}} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ f_0 & f_1 & f_2 \end{bmatrix} \\ \\ &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \end{aligned}$ $\begin{aligned} f_0(x, y, z) &= x\cos(z) \\ \\ f_1(x, y, z) &= y\sin(z) \\ \\ f_2(x, y, z) &= z\tan(z) \end{aligned}$ Let's calculate all the partial derivatives we'll need. $f_0$ $f_1$ $f_2$ $\dfrac{\partial}{\partial x}$ $0$ $0$ $\dfrac{\partial}{\partial y}$ $0$ $0$ $\dfrac{\partial}{\partial z}$ $-x\sin(z)$ $y\cos(z)$ Now we can put it all together. $\begin{aligned} \text{curl}(f) &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \\ \\ &= (0 - y\cos(z)) \hat{\imath} + (-x\sin(z) - 0) \hat{\jmath} + (0 - 0) \hat{k} \\ \\ &= -y\cos(z) \hat{\imath} - x\sin(z) \hat{\jmath} + 0 \hat{k} \end{aligned}$ In conclusion: $\text{curl}(f) = -y\cos(z) \hat{\imath} - x\sin(z) \hat{\jmath} + 0 \hat{k}$